\documentclass[reqno]{amsart}
\usepackage{hyperref}

\AtBeginDocument{{\noindent\small
{\em Electronic Journal of Differential Equations},
Vol. 2007(2007), No. 143, pp. 1--13.\newline
ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu
\newline ftp ejde.math.txstate.edu  (login: ftp)}
\thanks{\copyright 2007 Texas State University - San Marcos.}
\vspace{9mm}}

\begin{document}
\title[\hfilneg EJDE-2007/143\hfil Coexistence state]
{Coexistence state of a reaction-diffusion system}

\author[Y. Meng, Y. Wang \hfil EJDE-2007/143\hfilneg]
{Yijie Meng, Yifu Wang}  % in alphabetical order

\address{Yijie Meng\newline
Department of Mathematics, Xiang Fan University\\
Xiang Fan 441053, China}
\email{yijie\_meng@sina.com}

\address{Yifu Wang \newline
 Department of Applied Mathematics\\
  Beijing Institute of  Technology, Beijing 100081, China}
\email{yifu\_wang@163.com}

\thanks{Submitted May 10, 2007. Published October 25, 2007.}
\thanks{Supported  by grant 10401006 from the NSF of China}
\subjclass[2000]{35J55; 58J20} 
\keywords{Chemostat; competition model; principal eigenvalue; \hfill\break\indent
 maximum principle; Leray-Schauder degree}

\begin{abstract}
 Taking the spatial diffusion into account, we consider a
 reaction-diffusion system that models three species
 on a growth-limiting, nonreproducing resources in an unstirred
 chemostat. Sufficient conditions for the existence of a
 positive solution are determined. The main techniques is
 the  Leray-Schauder degree theory.
\end{abstract}

\maketitle
\numberwithin{equation}{section}
\newtheorem{theorem}{Theorem}[section]
\newtheorem{lemma}[theorem]{Lemma}
\newtheorem{remark}[theorem]{Remark}

\section{Introduction}

The chemostat is a laboratory apparatus used for the continuous
culture of microorganisms. It can be used to study competition
between different populations of microorganisms, and has the
advantage that the parameters are readily measurable. Experimental
verification of the match between theory and experiment in the
chemostat can be found in \cite{h1}. For a general discussion of
competition, see \cite{f1, s1}, while a detailed mathematical description
of competition in the chemostat can be found in \cite{s2}.

 In article \cite{b1}, a mathematical analysis  is given to a
competition model in a well-mixed chemostat with the equation the
\begin{equation}
\begin{gathered}
S'=(S^{0}-S)D-\frac{m_1S}{a_1+S}\frac{u_1}{\eta_1}
-\frac{m_2S}{a_2+S}\frac{u_2}{\eta_2},\\
u_1'=u_1(\frac{m_1S}{a_1+S}-D-\gamma u_3),\\
u_2'=u_2((1-k(u_1,u_2))\frac{m_2S}{a_2+S}-D),\\
u_3'=k(u_1,u_2)\frac{m_2Su_2}{a_2+S}-Du_3,
\end{gathered}\label{e1.1}
\end{equation}
where $S(t)$ denotes the nutrient concentration at time $t$,
$u_1(t)$ is the density of the sensitive microorganism at time
$t$, $u_2(t)$ is  the density of the toxin-producing organism at
time $t$, and $u_3(t)$ is the concentration of the toxicant at
time $t$, which is lethal to the microorganism $u_1(t)$. $S^{(0)}$
is the input concentration of nutrient, $D$ is the washout rate,
$m_i$ are the maximal growth rates, $a_i$ are the Michaelis-Menten
constants and $\eta_i$, $i=1,2$ are the yield constants. $S^{(0)}$
and $D$ are under the control of the experimenter, and the other
parameters are a function of the organism selected. The function
$k(u_1,u_2)$ represents the fraction of potential growth devoted
to producing the toxin and we assume that it is smooth. System
\eqref{e1.1} with constant $k$ is studied in \cite{h2} and it is noted
that  system \eqref{e1.1} is asymptotic to the standard
chemostat when $k\equiv 0$.
 The introduction of $k(u_1,u_2)$ requires
that a bacterium has the ability to sense the current state of its
habitat and the presence of other bacteria. The interaction
between the allelopathic agent and the sensitive microorganism has
been taken to be mass action form, $-\gamma u_3u_1$. This is
common modelling when an interaction depends on the two
concentrations.


 In the current paper, taking the spatial diffusion into
account, we remove the well-stirred hypothesis in system
\eqref{e1.1}, and thus are led to consider the following
reaction-diffusion (rescaled) system
\begin{equation}
\begin{gathered}
d_0\Delta S-m_1u_1f_1(S)-m_2u_2f_2(S)=0,\quad x\in\Omega\\
d_1\Delta u_1+m_1u_1f_1(S)-\gamma u_1u_3=0,\\
d_2\Delta u_2+m_2(1-k)u_2f_2(S)=0,\\
d_3\Delta u_3+m_2ku_2f_2(S)=0,
\end{gathered} \label{e1.2}
\end{equation}
subject to boundary conditions
\begin{equation}
\begin{gathered}
\frac{\partial S}{\partial n}+b(x)S=S^0(x),\quad
x\in\partial\Omega,\\
\frac{\partial u_i}{\partial n}+b(x)u_i=0(i=1,2,3),
\end{gathered}\label{e1.3}
\end{equation}
where $\Omega$ is a bounded domain in $R^N(N\geq 1)$ with smooth
boundary $\partial\Omega$, and $\frac{\partial}{\partial n}$
denotes the outward normal derivative, $0<k<1$.
$f_i(S)=\frac{S}{a_i+S}(i=1,2)$, $b(x)$ and $S^0(x)$ are
continuous on $\partial\Omega$, and $b(x),S^0(x)\geq 0,\not\equiv
0$, on $\partial\Omega$. $d_0$ is the diffusive coefficient for
the nutrient $S$, $d_i(i=1,2,3)$  are the random motility
coefficient of the microbial population $u_i$, respectively.

 Since only nonnegative solutions $(S,u_1,u_2,u_3)$ are
of biological interest, we redefine $\hat{f_i}(S)(i=1,2)$ for
$S<0$ as follows:
$$
\hat{f_i}(S)=\begin{cases}
f_i(S),& S\geq 0,\\
\arctan (\frac{2S}{a_i}+1)-\frac{\pi}{4},& S<0.
\end{cases}
$$
It is easily seen that $\hat{f_i}(S)\in C^1(R)$.


 The organization of this paper is as follows. In section
2, we obtain the existence and uniqueness of nonnegative
semi-trivial solutions by the principle eigenvalue problem and the
maximum principle. In section 3, the existence of  the positive
solution of  \eqref{e1.2}--\eqref{e1.3} is obtained by
making use of the theory of Leray-Schauder degree \cite{z1}.
\vspace{0.2cm}


\section{The Semi-trivial Solution}

In this section,  we shall consider the semi-trivial solutions of
 \eqref{e1.2}--\eqref{e1.3}. To this end, we first
investigate the following problem
\begin{equation}
\begin{gathered}
\Delta S=0,\quad  x\in\Omega,\\
\frac{\partial S}{\partial n}+b(x)S=S^0(x),\quad  x\in\partial\Omega,
\end{gathered} \label{e2.1}
\end{equation}

\begin{lemma} \label{lem2.1}
There exists a unique positive solution $S^*(x)$.
\end{lemma}

The proof can be seen in \cite{w1,z2}, and is omitted here.

From the maximum principle,  the nonnegative
solution $(S,u_1,u_2,u_3)$ of system
\eqref{e1.2}--\eqref{e1.3} satisfies
$$
d_0S+d_1u_1+d_2u_2+d_3u_3\leq d_0S^*(x),\quad x\in\Omega.
$$
Let $\lambda_i>0(i=1,2)$ be the principal eigenvalue of the
 problem
\begin{equation}
\begin{gathered}
d_i\Delta\phi+\lambda f_i(S^*)\phi=0,\quad  x\in\Omega,\\
\frac{\partial\phi}{\partial n}+b(x)\phi=0, \quad x\in\partial\Omega,
\end{gathered} \label{e2.2}
\end{equation}
with the corresponding eigenfunction $\phi_i>0(i=1,2)$.


\begin{theorem} \label{thm2.1}
If $m_1>\lambda_1$, then \eqref{e1.2}--\eqref{e1.3} admits a unique semi-trivial solution
$(S_1,\overline{u}_1,0,0)$ with
$d_0S_1+d_1\overline{u}_1=d_0S^*,S_1>0,\overline{u}_1>0$.
\end{theorem}

\begin{proof}
 Taking $u_2\equiv 0,u_3\equiv 0$, system
\eqref{e1.2}--\eqref{e1.3} is reduced to the  system
\begin{equation}
\begin{gathered}
d_0\Delta S-m_1u_1f_1(S)=0,\quad  x\in\Omega,\\
d_1\Delta u_1+m_1u_1f_1(S)=0,\\
\frac{\partial S}{\partial n}+b(x)S=S_0(x), \quad
x\in\partial\Omega,\\
\frac{\partial u_1}{\partial n}+b(x)u_1=0.
\end{gathered} \label{e2.3}
\end{equation}
Let $Z=d_0S+d_1u_1$, then $Z$ satisfies
\begin{equation}
\begin{gathered}
\Delta Z=0,\quad x\in\Omega,\\
\frac{\partial Z}{\partial n}+b(x)Z=d_0S_0(x), \quad
x\in\partial\Omega,
\end{gathered} \label{e2.4}
\end{equation}
from Lemma \ref{lem2.1}, \eqref{e2.4} have a unique positive solution
$S^*$, it satisfies $d_0S+d_1u_1=d_0S^*$. Thus $u_1$ satisfies
\begin{equation}
\begin{gathered}
d_1\Delta u_1+m_1u_1f_1(\frac{d_0S^*-d_1u_1}{d_0})=0,\quad  x\in\Omega,\\
\frac{\partial u_1}{\partial n}+b(x)u_1=0,\quad  x\in\partial\Omega,
\end{gathered} \label{e2.5}
\end{equation}
Arguing exactly as \cite[lemma 3.2]{w1}, \eqref{e2.5} have a unique
positive solution $\overline{u}_1$. Thus we complete the proof of
the theorem.
\end{proof}

 Similar to the proof of \cite[Lemma 3.2]{w1}, we can also
show that the  system
\begin{equation}
\begin{gathered}
d_2\Delta\psi+m_2(1-k)\psi f_2(S^*-\frac{d_2\psi}{d_0})=0,\quad x\in\Omega,\\
\frac{\partial\psi}{\partial n}+b(x)\psi=0,\quad  x\in\partial\Omega
\end{gathered}\label{e2.6}
\end{equation}
has a unique positive solution $\psi<\frac{d_0}{d_2}S^*$ if
$m_2>\frac{\lambda_2}{1-k}$.

\begin{theorem} \label{thm2.2}
If $m_2>\frac{\lambda_2}{1-k}$, then system
\eqref{e1.2}--\eqref{e1.3} admits a unique semi-trivial solution
$(S_2,0,\overline{u}_2,\overline{u}_3)$ with
$d_0S_2+d_2\overline{u}_2+d_3\overline{u}_3=d_0S^*$,
$S_2>0$, $\overline{u}_2>0$,
$\overline{u}_3>0$.
\end{theorem}

 \begin{proof} We take $u_1\equiv 0$ and thus reduce
 \eqref{e1.2}--\eqref{e1.3} to the system
\begin{equation}
\begin{gathered}
d_0\Delta S-m_2u_2f_2(S)=0,\quad x\in\Omega\\
d_2\Delta u_2+m_2(1-k)u_2f_2(S)=0,\\
d_3\Delta u_3+m_2ku_2f_2(S)=0,\\
\frac{\partial S}{\partial n}+b(x)S=S^0(x), \quad
x\in\partial\Omega,\\
\frac{\partial u_i}{\partial n}+b(x)u_i=0(i=2,3),
\end{gathered} \label{e2.7}
\end{equation}
Let $z=d_0S+d_2u_2+d_3u_3$, then $\frac{z}{d_0}$ satisfies
\eqref{e2.1}, and thus $z=d_0S^*(x)$. Substituting it into
\eqref{e2.7}, we get the reduced boundary-value problem
\begin{equation}
\begin{gathered}
d_2\Delta u_2+m_2(1-k)u_2f_2(S^*-\frac{d_2u_2+d_3u_3}{d_0})=0,\quad
 x\in\Omega,\\
d_3\Delta u_3+m_2ku_2f_2(S^*-\frac{d_2u_2+d_3u_3}{d_0})=0,\\
\frac{\partial u_i}{\partial n}+b(x)u_i=0(i=2,3), \quad
x\in\partial\Omega.
\end{gathered} \label{e2.8}
\end{equation}
It is easy to check that $((1-k)\psi,\frac{d_2}{d_3}k\psi)$ is
exactly a positive solution of \eqref{e2.8}.

Let $(u_2,u_3)$ be the positive solution of \eqref{e2.8}, and
$w=d_2ku_2-d_3(1-k)u_3$, then $w$ satisfies
\begin{gather*}
\Delta w=0,\quad  x\in\Omega,\\
\frac{\partial w}{\partial n}+b(x)w=0, \quad x\in\partial\Omega.
\end{gather*}
By the maximum principle, we get $w=0$ and $d_2ku_2=d_3(1-k)u_3$.

  Substituting $d_2ku_2=d_3(1-k)u_3$ into \eqref{e2.8}, we have
\begin{gather*}
d_2\Delta u_2+m_2(1-k)u_2 f_2(S^*-\frac{d_2u_2}{d_0(1-k)})=0,\\
\frac{\partial u_2}{\partial n}+b(x)u_2=0.
\end{gather*}
Then by the uniqueness of the positive solution of \eqref{e2.6},
it follows that $u_2=(1-k)\psi$ and thus
$u_3=\frac{d_2}{d_3}k\psi$.  The proof is complete.
\end{proof}

At this position, we can get the following result which implies
that $m_1>\lambda_1,m_2>\frac{\lambda_2}{1-k}$ is necessary to the
existence of coexistence states of system \eqref{e1.2}--\eqref{e1.3}.

\begin{theorem} \label{thm2.3}
If $m_1\leq\lambda_1$ or
$m_2\leq\frac{\lambda_2}{1-k}$ are satisfied, then the nonnegative solution
$(S,u_1,u_2,u_3)$ of  system \eqref{e1.2}--\eqref{e1.3}
satisfies $u_1=0$ or
$u_2=u_3=0$.
\end{theorem}

The proof is omitted here and the reader may refer to
\cite{h3,h4,w1}. Therefore we suppose
$m_1>\lambda_1,m_2>\frac{\lambda_2}{1-k}$ in what follows.


\section{Coexistence State}
   In this section, following some ideas of Dung and Smith \cite{d3},
we shall determine the
sufficient conditions for the existence of positive solutions of
\eqref{e1.2}--\eqref{e1.3} by the theory of
Leray-Schauder degree .
Now we state some results about Leray-Schauder degree, which
appeared in \cite{a1,z1}.

\begin{theorem} \label{thmA}
Let $X$ be a retract of some Banach space $E$,
let $U$ be an open subset of $X$, and let
$A:\overline{U}\to X$ be a compact map. Suppose that
$x_0\in U$ is a fixed point of $A$, and suppose that there exists
a positive number $\rho$ such that $x_0+\rho B\subset U$, where
$B$ denotes the open unit ball of $E$. Finally, suppose that $A$
is differentiable at $x_0$, such that $1$ is not an eigenvalue of
the derivative $A'(x_0)$. Then $x_0$ is an isolated fixed point
of $A$, and
$$
\mathop{\rm index}(x_0,A)=\deg(I-A,B,0)=\deg(I-A'(x_0),
\overline{B},0)=(-1)^m,
$$
where $m$ is the sum of the multiplicities of all the eigenvalues
of $A'(x_0)$ which are greater than one.
\end{theorem}

For every $\rho>0$, we denote by $P_{\rho}$ the positive part of
$\rho B$, that is $P_{\rho}=\rho B\cap P=\rho B^+$.

\begin{lemma} \label{lemA}
 Let $A:\overline{P}_{\rho}\to P$ be a
compact map such that $A(0)=0$. Suppose that $A$ has a right
derivative $A'_{+}(0)$ at zero such that $1$ is not an
eigenvalue of $A'_{+}(0)$ to a positive eigenvector. Then there
exists a constant $\sigma\in(0,\sigma_0]$,
\begin{itemize}
\item[(i)] $\mathop{\rm index}(P_{\rho},A)=\deg(I-A,P_{\rho},0)=1$
if $A'_{+}(0)$ has no positive eigenvector to an eigenvalue greater than one;

\item[(ii)] $\mathop{\rm index}(P_{\rho},A)=0$ if $A'_{+}(0)$ possesses
a positive eigenvector to an eigenvalue greater than one.
\end{itemize}
\end{lemma}

Let
\begin{gather*}
C_0(\overline{\Omega})=\{\phi\in
C(\overline{\Omega}):\frac{\partial\phi}{\partial n
}+b(x)\phi=0\quad \hbox{on }\partial\Omega\}\\
K=\{\phi\in C_0(\overline{\Omega}):\phi\geq 0\quad \hbox{in }\Omega\},
\end{gather*}
then $C_0(\overline{\Omega})$ is a Banach space and $K$ is a
positive cone in $C_0(\overline{\Omega})$. We define
$$
E=C_0(\overline{\Omega})\times C_0(\overline{\Omega})\times
C_0(\overline{\Omega})\times
C_0(\overline{\Omega}),\quad E_+=K\times K\times K\times K.
$$
Let $s(x)=S^*(x)-S(x)$, then system \eqref{e1.2}--\eqref{e1.3}
becomes
\begin{equation}
\begin{gathered}
d_0\Delta s+m_1u_1f_1(S^*-s)+m_2u_2f_2(S^*-s)=0,\quad  x\in\Omega,\\
d_1\Delta u_1+m_1u_1f_1(S^*-s)-\gamma u_1u_3=0,\\
d_2\Delta u_2+m_2(1-k)u_2f_2(S^*-s)=0,\\
d_3\Delta u_3+m_2ku_2f_2(S^*-s)=0,\\
\frac{\partial s}{\partial n}+b(x)s=0, \quad x\in\partial\Omega,\\
\frac{\partial u_i}{\partial n}+b(x)u_i=0(i=1,2,3).
\end{gathered} \label{e3.1}
\end{equation}
Since $S(x)\geq 0$, the nonnegative solution of \eqref{e3.1}
satisfies  $s(x)\leq S^*(x)$.
Define $A:E_+\to E$ by
\begin{align*}
A(s,u_1,u_2,u_3)
&=\Big((-d_0\Delta)^{-1}(m_1u_1f_1(S^*-s)+m_2u_2f_2(S^*-s)),\\
&\quad (-d_1\Delta)^{-1}(m_1u_1f_1(S^*-s)-\gamma u_1u_3),\\
&\quad (-d_2\Delta)^{-1}(m_2(1-k)u_2f_2(S^*-s)),\\
&\quad (-d_3\Delta)^{-1}(m_2ku_2f_2(S^*-s))\Big)
\end{align*}
for $(s,u_1,u_2,u_3)\in E_+$, then it is observed that $A$ is a
compact operator and every nonnegative  solution of
\eqref{e1.2}--\eqref{e1.3} corresponds to the fixed point of the
operator $A$ on the cone $E_+$.

Clearly, $e_0=(0,0,0,0)$,
$e_1=(S^*-S_1,\overline{u}_1,0,0)$,
$e_2=(S^*-S_2,0,\overline{u}_2,\overline{u}_3)$
are fixed points of compact operator $A$.

Let $\lambda'_i(q)(i=1,2)$ be the principal eigenvalue of
$$
-d_i\Delta\omega+q(x)\omega=\lambda\omega,
$$
with $q(x)\in C(\overline{\Omega})$. It is well known \cite{d2} that
$\lambda'_i(q)$ depends continuously on $q$, and $q_1\leq
q_2,q_1\not\equiv q_2$ implies $\lambda'_i(q_1)<\lambda'_i(q_2)$.
 From \cite[Theorem 4.3, 4.4 and 4.5]{a1},
$\lambda'_1(-m_1f_1(S^*))<0, \lambda'_2(-m_2(1-k)f_2(S^*))<0$ is
equivalent to $m_1>\lambda_1, m_2>\frac{\lambda_2}{(1-k)}$,
respectively.


\begin{lemma} \label{lem3.1}
$\mathop{\rm index}(e_0,A)=0$.
\end{lemma}

\begin{proof} Let $A'(e_0)$ be the derivative of $A$ at $e_0$.
If $\lambda>0$ is an eigenvalue of $A'(e_0)$ corresponding to the
eigenfunction $(s,u_1,u_2,u_3)^{T}\in E_+$. Then
\begin{gather*}
d_0\Delta s+\frac{1}{\lambda}(m_1u_1f_1(S^*)+m_2u_2f_2(S^*))=0,\quad x\in\Omega\\
d_1\Delta u_1+\frac{1}{\lambda}m_1u_1f_1(S^*)=0,\\
d_2\Delta u_2+\frac{1}{\lambda}m_2(1-k)u_2f_2(S^*)=0,\\
d_3\Delta u_3+\frac{1}{\lambda}m_2ku_2f_2(S^*)=0,
\end{gather*}
with homogeneous boundary conditions.
 From $\lambda'_1(-m_1f_1(S^*))<0$ and
$\lambda'_2(-m_2(1-k)f_2(S^*))<0$, we have  $\lambda\neq 1$.
 From \eqref{e2.2} and $m_1>\lambda_1$, it follows that
$$
d_1\Delta \phi_1+\frac{1}{\lambda_0}m_1\phi_1f_1(S^*)=0
$$
with $\lambda_0=\frac{m_1}{\lambda_1}>1$. Then
\begin{align*}
&A'(e_0)((-d_0\Delta)^{-1}\frac{1}{\lambda_0} m_1\phi_1f_1(S^*),\phi_1,0,0)^T \\
 &= \lambda_0 ((-d_0\Delta)^{-1}\frac{1}{\lambda_0} m_1\phi_1f_1(S^*),
 \phi_1,0,0)^T.
\end{align*}
Thus  from Lemma \ref{lemA}, it follows that $\mathop{\rm index}(e_0,A)=0$.
\end{proof}

\begin{lemma} \label{lem3.2}
 There exists a constant $R>0$ such that
$\deg(I-A,P_R,0)=1$, where $P_R=\{U\in E_+:\|s\|<R,\| u_i\|<R\}(i=1,2,3)$.
\end{lemma}

\begin{proof}
For $t\in[0,1]$, $(s,u_1,u_2,u_3)=tA(s,u_1,u_2,u_3)$ and
$(s,u_1,u_2,u_3)\in E_+$
implies
\begin{equation}
\begin{gathered}
d_0\Delta s+tm_1u_1f_1(S^*-s)+tm_2u_2f_2(S^*-s)=0,\quad x\in\Omega,\\
d_1\Delta u_1+tm_1u_1f_1(S^*-s)-t\gamma u_1u_3=0,\\
d_2\Delta u_2+tm_2(1-k)u_2f_2(S^*-s)=0,\\
d_3\Delta u_3+tm_2ku_2f_2(S^*-s)=0,\\
\frac{\partial s}{\partial n}+b(x)s=0, \quad x\in\partial\Omega,\\
\frac{\partial u_i}{\partial n}+b(x)u_i=0(i=1,2,3).
\end{gathered} \label{e3.2}
\end{equation}
We claim that $s\leq S^*$. Otherwise, if $g(x)=s(x)-S^*(x)$
attains its maximum at some point $x_0\in\overline\Omega$, then
$g(x_0)>0$ and $g(x)$ satisfies
$$
d_0\Delta g+tm_1u_1f_1(-g)+tm_2u_2f_2(-g)=0.\label{e3.3}
$$
Now if $u_1=0$ and $u_2=0$ or $t=0$, then from the first equation
in \eqref{e3.2}, it follows $s\equiv 0$ and thus $s\leq S^*$
holds. So we assume that $t>0$ and  $u_1\not\equiv 0$ or
$u_2\not\equiv 0$. By the maximum principle, $u_1>0$ or $u_2>0$.

If $x_0\in\Omega$, then $\Delta g(x_0)\leq 0$. However, $g(x_0)>0$
implies
\[
d_0\Delta g(x_0)=tm_1u_1(x_0)f_1(-g(x_0))-tm_2u_2(x_0)f_2(-g(x_0))>0,
\]
a contradiction. Hence $x_0\in\partial\Omega$ and thus
$\frac{\partial g}{\partial n}|_{x_0}>0$ by the maximum principle.
On the other hand,
$\frac{\partial g}{\partial n}|_{x_0}+b(x_0)g(x_0)=-S^0(x_0)\leq 0$
implies that $\frac{\partial g}{\partial n}|_{x_0}\leq 0$, a contradiction.
Therefore $s\leq S^*$ on $\overline\Omega$.
Let $\hat{S}=S^*-s$, then  system \eqref{e3.2} becomes
\begin{gather*}
d_0\Delta\hat{S}-tm_1u_1f_1(\hat{S})-tm_2u_2f_2(\hat{S})=0,\quad x\in\Omega,\\
d_1\Delta u_1+tm_1u_1f_1(\hat{S})-t\gamma u_1u_3=0,\\
d_2\Delta u_2+tm_2(1-k)u_2f_2(\hat{S})=0,\\
d_3\Delta u_3+tm_2ku_2f_2(\hat{S})=0,\\
\frac{\partial\hat{S}}{\partial n}+b(x)\hat{S}=S^0(x),\quad
x\in\partial\Omega,\\
\frac{\partial u_i}{\partial n}+b(x)u_i=0(i=1,2,3),
\end{gather*}
It follows that $d_0\hat{S}+d_1u_1+d_2u_2+d_3u_3\leq d_0S^*$.
Hence $u_i\leq \frac{d_0S^*}{d_i}(i=1,2,3)$. Let
$M=\max\{1,\frac{d_0}{d_1},\frac{d_0}{d_2},\frac{d_0}{d_3}\}$,
$R=M\max_{x\in\overline{\Omega}}S^*(x)$, Then
$U=(s,u_1,u_2,u_3)\not\in\partial P_R$. By the homotopy invariance
of the degree, we obtain
$$
\deg(I-A,P_R,0)=\deg(I,P_R,0)=1.
$$
\end{proof}

\begin{lemma} \label{lem3.3}
Suppose $\lambda'_2(-m_2(1-k)f_2(S_1))<0$, then
$\mathop{\rm index}(e_1,A)=0$.
\end{lemma}

\begin{proof}
Consider problem \eqref{e3.1} in the form
\begin{equation}
\begin{gathered}
d_0\Delta s+m_1u_1f_1(S^*-s)+tm_2u_2f_2(S^*-s)=0,\quad x\in\Omega\\
d_1\Delta u_1+m_1u_1f_1(S^*-s)-t\gamma u_1u_3=0,\\
d_2\Delta u_2+m_2(1-k)u_2f_2(S^*-s)=0,\\
d_3\Delta u_3+m_2ku_2f_2(S^*-s)=0,\\
\frac{\partial s}{\partial n}+b(x)s=0, \quad
x\in\partial\Omega,\\
\frac{\partial u_i}{\partial n}+b(x)u_i=0(i=1,2,3),
\end{gathered} \label{e3.4}
\end{equation}
where the parameter $t=1$.

Here we regard $t\in [0,1]$ as the homotopy parameter and hence
equivalent fixed point problem can be denoted by $U=H(t,U)$. It is
obvious that $H(1,U)=A(U)$.

We assume that $A(U)=U$ has no one positive solution in
$P_R\setminus\overline{P_r}(r<<1)$, otherwise there are nothing to
do.

Choose a neighborhood $Q=V\times W$of $e_1$ in
$P_R\setminus\overline{P_r}$, where $V$ is a neighborhood
$(S^*-S_1,\overline{u}_1)$ in $C_0(\overline{\Omega})\times
C_0(\overline{\Omega})$, and $W$ is a small neighborhood of
$(0,0)$ in $C_0(\overline{\Omega})\times C_0(\overline{\Omega})$.

If $H(0,U)=U$ has a solution $U=(s,u_1,u_2,u_3)\in\partial Q$,
which implies $u_1\neq 0$, then $(s,u_1)=(S^*-S_1,\overline{u}_1)$
by Theorem \ref{thm2.1}. If $u_2=0$, then $U=e_1$, but $e_1\not\in\partial
Q$. Therefore $u_2>0$ and thus we have a contradiction to
$\lambda'_2(-m_2(1-k)f_2(S_1))<0$.

If there exists $t\in (0,1]$ such that $H(t,U)=U$ has a solution
$U=(s,u_1,u_2,u_3)\in\partial Q$, then $u_2\neq 0$. Since once
$u_2\equiv 0$, then $U=e_1$ contradicting $U\in\partial Q$.
Therefore $u_2>0$, and thus $(s,u_1,tu_2,tu_3)$ is a positive
fixed point of $A$ contradicting our assumption.

By the homotopy invariance of Leray-Schauder degree
$$
\mathop{\rm index}(e_1,A)=\mathop{\rm index}(e_1,H(1,\cdot))
=\mathop{\rm index}(e_1,H(0,\cdot)).
$$
Now consider the boundary-value problem with parameter $t\in
[0,1]$
\begin{equation}
\begin{gathered}
d_0\Delta s+m_1u_1f_1(S^*-s)=0,\quad x\in\Omega,\\
d_1\Delta u_1+m_1u_1f_1(S^*-s)=0,\\
d_2\Delta u_2+m_2(1-k)u_2f_2(tS_1+(1-t)(S^*-s))=0,\\
d_3\Delta u_3+m_2ku_2f_2(tS_1+(1-t)(S^*-s))=0,\\
\frac{\partial s}{\partial n}+b(x)s=0,\quad
x\in\partial\Omega,\\
\frac{\partial u_i}{\partial n}+b(x)u_i=0(i=1,2,3).
\end{gathered} \label{e3.5}
\end{equation}
In fixed point form, system \eqref{e3.5} becomes  $G(t,U)=U$.
If $G(t,U)=U$ for some $t\in [0,1]$ and
$U=(s,u_1,u_2,u_3)\in\partial Q$, then obviously
$(s,u_1)=(S^*-S_1,\overline{u}_1)$, and so $u_2\equiv 0$ by
$\lambda'_2(-m_2(1-k)f_2(S_1))<0$. Thus $U=e_1$ contradicting
$e_1\not\in\partial Q$. Again, by the homotopy invariance of
Leray-Schauder degree,
$$
\mathop{\rm index}(e_1,A)=\mathop{\rm index}(e_1,H(0,\cdot))
=\mathop{\rm index}(e_1,G(0,\cdot))=\mathop{\rm index}(e_1,G(1,\cdot)).
$$
However, $G(1,\cdot)$ can be view as the product of two maps $G_1$
on $V$ and $G_2$ on $W$, which are associated with the boundary
value problems
\begin{gather*}
d_0\Delta s+m_1u_1f_1(S^*-s)=0,\\
d_1\Delta u_1+m_1u_1f_1(S^*-s)=0,
\end{gather*}
and
\begin{gather*}
d_2\Delta u_2+m_2(1-k)u_2f_2(S_1)=0,\\
d_3\Delta u_3+m_2ku_2f_2(S_1)=0,
\end{gather*}
with homogeneous boundary conditions, respectively.

Now, by the uniqueness of $(S^*-S_1,\overline{u}_1)$ and
$m_1>\lambda_1$,
$$
\deg(G_1,V,(0,0))=\mathop{\rm index}((S^*-S_1,\overline{u}_1),G_1)=1.
$$
Furthermore from $\lambda'_2(-m_2(1-k)f_2(S_1))<0$, it follows
$$
 \deg(G_2,W,(0,0))=\mathop{\rm index}((0,0),G_2).
$$
In fact, if $\lambda>0$ is an eigenvalue of $G'_2(0,0)=G_2$
corresponding to the eigenfunction $(u_2,u_3)^T\in W$, then
\begin{gather*}
d_2\Delta u_2+\frac{1}{\lambda}m_2(1-k)u_2f_2(S_1)=0,\\
d_3\Delta u_3+\frac{1}{\lambda}m_2ku_2f_2(S_1)=0.
\end{gather*}
 By $\lambda'_2(-m_2(1-k)f_2(S_1))<0$, $\lambda\neq 1$. Therefore
there exists $\lambda>1$ and $u_2>0$ the corresponding
eigenfunction such that
$$
d_2\Delta u_2+\frac{1}{\lambda}m_2(1-k)u_2f_2(S_1)=0.
$$
Thus
$G'_2(0,0)(u_2,(-d_3\Delta)^{-1}(\frac{1}{\lambda}m_2kf_2(S_1)))^T
=\lambda(u_2,(-d_3\Delta)^{-1}(\frac{1}{\lambda}m_2kf_2(S_1)))^T$.
It follows from Lemma \ref{lemA} that $\mathop{\rm index}((0,0),G_2)=0$.

By the product theorem of Leray-Schauder degree
\cite[Theorem 13.F]{z1}
$$
\mathop{\rm index}(e_1,A)=\deg(G_1,V,(0,0))\deg(G_2,W,(0,0))=0.
$$
\end{proof}

\begin{lemma} \label{lem3.4}
Suppose $\lambda'_1(-m_1f_1(S_2)+\gamma\overline{u}_3)<0$, then
$\mathop{\rm index}(e_2,A)$=0.
\end{lemma}

\begin{proof} Consider \eqref{e3.1} in the form
\begin{equation}
\begin{gathered}
d_0\Delta s+tm_1u_1f_1(S^*-s)+m_2u_2f_2(S^*-s)=0,\quad  x\in\Omega\\
d_1\Delta u_1+m_1u_1f_1(S^*-s)-\gamma u_1u_3=0,\\
d_2\Delta u_2+m_2(1-k)u_2f_2(S^*-s)=0,\\
d_3\Delta u_3+m_2ku_2f_2(S^*-s)=0,\\
\frac{\partial s}{\partial n}+b(x)s=0, \quad
x\in\partial\Omega,\\
\frac{\partial u_i}{\partial n}+b(x)u_i=0(i=1,2,3).
\end{gathered} \label{e3.6}
\end{equation}
with the parameter $t=1$.

Here we regard $t\in [0,1]$ as the homotopy parameter and hence
equivalent fixed point problem can be denoted by $U=H(t,U)$. It is
obvious that $H(1,U)=A(U)$.

We assume that $A(U)=U$ has no one positive solution in
$P_R\setminus\overline{P_r}(r<<1)$, otherwise there are nothing to
do.

Choose a neighborhood $Q=V\times W$ of $e_2$ in
$P_R\setminus\overline{P_r}$, where $V$ is a neighborhood
$(S^*-S_2,\overline{u}_2,\overline{u}_3)$ in
$C_0(\overline{\Omega})\times C_0(\overline{\Omega})\times
C_0(\overline{\Omega})$, and $W$ is a small neighborhood of $(0)$
in $C_0(\overline{\Omega})$.

If $H(0,U)=U$ has a solution $U=(s,u_1,u_2,u_3)\in\partial Q$,
which implies $u_2\neq 0$, then
$(s,u_2,u_3)=(S^*-S_2,\overline{u}_2,\overline{u}_3)$ by
Theorem \ref{thm2.2}. If $u_1=0$, then $U=e_2$, but $e_2\not\in\partial Q$.
Therefore $u_1>0$ and thus we have a contradiction to
$\lambda'_1(-m_1f_1(S_2)+\gamma\overline{u}_3)<0$.

If there exists $t\in (0,1]$ such that $H(t,U)=U$ has a solution
$U=(s,u_1,u_2,u_3)\in\partial Q$, then $u_1\neq 0$. Since once
$u_1\equiv 0$, then $U=e_2$ contradicting $U\in\partial Q$.
Therefore $u_1>0$, and thus $(s,tu_1,u_2,u_3)$ is a positive fixed
point of $A$ contradicting our assumption.

By the homotopy invariance of Leray-Schauder degree
$$
\mathop{\rm index}(e_2,A)=\mathop{\rm index}(e_2,H(1,\cdot))
=\mathop{\rm index}(e_2,H(0,\cdot)).
$$
Now consider the boundary value problem with parameter $t\in
[0,1]$
\begin{equation}
\begin{gathered}
d_0\Delta s+m_2u_2f_2(S^*-s)=0,\quad  x\in\Omega,\\
d_1\Delta u_1+m_1u_1f_1(tS_2+(1-t)(S^*-s))-\gamma u_1u_3=0,\\
d_2\Delta u_2+m_2(1-k)u_2f_2(S^*-s)=0,\\
d_3\Delta u_3+m_2ku_2f_2(S^*-s)=0,\\
\frac{\partial s}{\partial n}+b(x)s=0, \quad
x\in\partial\Omega,\\
\frac{\partial u_i}{\partial n}+b(x)u_i=0(i=1,2,3).
\end{gathered}\label{e3.7}
\end{equation}
In fixed point form,  system \eqref{e3.7} becomes $G(t,U)=U$.
If $G(t,U)=U$ for some $t\in [0,1]$ and
$U=(s,u_1,u_2,u_3)\in\partial Q$, then obviously
$(s,u_2,u_3)=(S^*-S_2,\overline{u}_2,\overline{u}_3)$, and so
$u_1\equiv 0$ by
$\lambda'_1(-m_1f_2(S_1)+\gamma\overline{u}_3)<0$. Thus $U=e_2$
contradicting $e_2\not\in\partial Q$. Again, by the homotopy
invariance of Leray-Schauder degree,
$$
\mathop{\rm index}(e_2,A)=\mathop{\rm index}(e_2,H(0,\cdot))
=\mathop{\rm index}(e_2,G(0,\cdot))=\mathop{\rm index}(e_2,G(1,\cdot)).
$$
Next, consider the boundary value problem with parameter $t\in
[0,1]$
\begin{equation}
\begin{gathered}
d_0\Delta s+m_2u_2f_2(S^*-s)=0,\quad x\in\Omega,\\
d_1\Delta u_1+m_1u_1f_1(S_2)-t\gamma u_1u_3=0,\\
d_2\Delta u_2+m_2(1-k)u_2f_2(S^*-s)=0,\\
d_3\Delta u_3+m_2ku_2f_2(S^*-s)=0,\\
\frac{\partial s}{\partial n}+b(x)s=0, \quad
x\in\partial\Omega,\\
\frac{\partial u_i}{\partial n}+b(x)u_i=0(i=1,2,3).
\end{gathered}\label{e3.8}
\end{equation}
In fixed point form,  system \eqref{e3.8} becomes $K(t,U)=U$.
If $K(t,U)=U$ for some $t\in [0,1]$ and
$U=(s,u_1,u_2,u_3)\in\partial Q$, then obviously
$(s,u_2,u_3)=(S^*-S_2,\overline{u}_2,\overline{u}_3)$, and so
$u_1\equiv 0$ by
$\lambda'_1(-m_1f_1(S_2)+\gamma\overline{u}_3)<0$. Thus $U=e_2$
contradicting $e_2\not\in\partial Q$. Again, by the homotopy
invariance of Leray-Schauder degree,
$$
\mathop{\rm index}(e_2,A)=\mathop{\rm index}(e_2,G(1,\cdot))
=\mathop{\rm index}(e_2,K(1,\cdot))=\mathop{\rm index}(e_2,K(0,\cdot)).
$$
However, $K(0,\cdot)$ can be view as the product of two maps $K_1$
on $V$ and $K_2$ on $W$, which are associated with the boundary
value problems
\begin{gather*}
d_0\Delta s+m_2u_2f_2(S^*-s)=0,\\
d_2\Delta u_2+m_2(1-k)u_2f_2(S^*-s)=0,\\
d_3\Delta u_3+m_2ku_2f_2(S^*-s)=0,\\
d_1\Delta u_1+m_1u_1f_1(S_2)=0,
\end{gather*}
with  homogeneous boundary conditions, respectively.

Now, by the uniqueness of
$(S^*-S_2,\overline{u}_2,\overline{u}_3)$ and
$m_2>\lambda_2/(1-k)$,
$$
\deg(K_1,V,0)=\mathop{\rm index}((S^*-S_2,\overline{u}_2,\overline{u}_3)
,K_1)=1.
$$
Furthermore from
$\lambda'_1(-m_1f_1(S_2))<\lambda'_1(-m_1f_1(S_2)+\gamma\overline{u}_3)<0$,
it follows that
$$
 \deg(K_2,W,0)=\mathop{\rm index}(0,K_2)=0.
$$
In fact, if $\lambda>0$ is an eigenvalue of $K'_2(0)=K_2$
corresponding to the eigenfunction $u_1\in W$, then
$$
d_1\Delta u_1+\frac{1}{\lambda}m_1u_1f_1(S_2)=0.
$$
By $\lambda'_1(-m_1f_1(S_2)+\gamma\overline{u}_3)<0$, $\lambda\neq
1$. Therefore there exist $\lambda>1$ and the corresponding
eigenfunction $u_1>0$ such that
$$
d_1\Delta u_1+\frac{1}{\lambda}m_1u_1f_1(S_2)=0.
$$
It follows from Lemma \ref{lemA} that $\mathop{\rm index}(0,K_2)=0$.

By the product theorem of Leray-Schauder degree \cite{z1},
$$
\mathop{\rm index}(e_2,A)=\deg(K_1,V,,0))\deg(K_2,W,0)=0.
$$
\end{proof}

Therefore, by the additivity property of the fixed point index and
above Lemmas, we have the following result.

\begin{theorem} \label{thm3.1}
Assume that $\lambda'_1(-m_1f_1(S_2)+\gamma\overline{u}_3)<0$ and
$\lambda'_2(-m_2(1-k)f_2(S_1))<0$, then system
\eqref{e1.2}--\eqref{e1.3} admits at least one positive solution.
\end{theorem}

 We note that $\lambda'_1(-m_1f_1(S_2)+\gamma\overline{u}_3)<0$ and
$\lambda'_2(-m_2(1-k)f_2(S_1))<0$ implies
$\lambda'_1(-m_1f_1(S^*)<0$ and $\lambda'_2(-m_2(1-k)f_2(S^*))<0$
respectively, since $S_1, S_2\leq S^*$ and the monotonicity of
function $f_i$.

For the other case, we present the following results, whose
proofs are very similar to that of Lemmas \ref{lem3.3},
\ref{lem3.4}  and Theorem \ref{thm3.1}.


\begin{lemma} \label{lem3.5}
Assume that $\lambda'_2(-m_2(1-k)f_2(S_1))>0$,
then $\mathop{\rm index}(e_1,A)=1$.
\end{lemma}

\begin{lemma} \label{lem3.6}
Assume that $\lambda'_1(-m_1f_1(S_2)+\gamma\overline{u}_3)>0$, then
$\mathop{\rm index}(e_2,A)=1$.
\end{lemma}

\begin{theorem} \label{thm3.2}
 Assume that $\lambda'_1(-m_1f_1(S_2)+\gamma\overline{u}_3)>0$ and
that
\[
\lambda'_2(-m_2(1-k)f_2(S_1))>0,
\]
 then \eqref{e1.2}--\eqref{e1.3} admits at least one positive solution.
\end{theorem}

\begin{remark} \rm
(1) If $(s,u_1,u_2,u_3)$ is the positive solution of
  \eqref{e1.2}--\eqref{e1.3}, then $u_i\leq\overline{u}_i(i=1,2,3)$
 by the maximum  principle. \smallskip

\noindent (2)  Our results implies that the existence of positive steady sates
of  \eqref{e1.2}--\eqref{e1.3} if the semi-trivial
nonegative solutions are stable or unstable simultaneously. \smallskip

\noindent (3) System \eqref{e1.2}--\eqref{e1.3} with $\gamma=0$ is
fundamentally   more tractable than the general case and rather complete
analysis can be done, due to the existence of a ``conservation principle"
which allows the reduction of   system \eqref{e1.1}-\eqref{e1.2}
to the competition system. In fact, if $\gamma=0$, then
 $d_0S+d_1u_1+d_2u_2+d_3u_3=d_0S^*$,
and thus system \eqref{e1.2}--\eqref{e1.3} reduces to the
competition system
\begin{equation}
\begin{gathered}
d_1\Delta u_1+
m_1u_1f_1(S^*-\frac{d_1u_1+\frac{d_2}{1-k}u_2}{d_0})=0,\quad x\in\Omega,\\
d_2\Delta u_2+
m_2(1-k)u_2f_2(S^*-\frac{d_1u_1+\frac{d_2}{1-k}u_2}{d_0})=0,\\
\frac{\partial u_i}{\partial n}+b(x)u_i=0(i=1,2), \quad
x\in\partial\Omega,
\end{gathered} \label{e3.9}
\end{equation}
noticing $d_2ku_2=d_3(1-k)u_3$. By the above results,
\eqref{e3.9} has at least one positive coexistence solution
$(u_1,u_2)$ if
$\lambda'_1(-m_1f_1(S_2))\cdot\lambda'_2(-m_2(1-k)f_2(S_1))>0$ .
Now we assume that $\lambda'_1(-m_1f_1(S_2))<0$ and
$\lambda'_2(-m_2(1-k)f_2(S_1))>0$.

 We define $u_1^n$ to be the
unique nonnegative nontrivial solution of
$$
d_1\Delta
u_1+m_1u_1f_1(S^*-\frac{d_1u_1+\frac{d_2}{1-k}u_2^{n-1}}{d_0})=0
$$
and $u_2^n$ to be the unique nonnegative nontrivial solution of
$$
d_2\Delta
u_2+m_2(1-k)u_2f_2(S^*-\frac{d_1u_1^n+\frac{d_2}{1-k}u_2}{d_0})=0,
$$
with $u_2^0=\overline{u}_2$, respectively. Thus
$u_1^1<u_1^2<\dots<u_1^n<\dots$ and
$u_2^1>u_2^2>\dots>u_2^n>\dots$. By arguments
in \cite{d1}, we can conclude that if  $\lambda'_1(-m_1f_1(S_2))<0$ and
$\lambda'_2(-m_2(1-k)f_2(S_1))>0$, then  system \eqref{e3.9}
has the coexistence solutions if and only if
$\lambda'_2(-m_2(1-k)f_2(S^*-\frac{d_1u_1^n}{d_0}))<0$, for all
$n\in N$. A similar result holds for $\lambda'_1(-m_1f_1(S_2))>0$
and $\lambda'_2(-m_2(1-k)f_2(S_1))<0$.
\end{remark}


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\end{document}